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Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 29 maths

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Answer:

2(\sqrt{2}-1)

Hint:

First integrate and then use limits

Given:

\int_{0}^{\frac{\pi}{2}} \sqrt{1-\sin 2 x} \; d x

Solution:

\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \sqrt{1-\sin 2 x} d x \\\\ &I=\int_{0}^{\frac{\pi}{2}} \sqrt{|\sin x-\cos x|^{2}} d x \end{aligned}

\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}}|\sin x-\cos x| d x \\\\ &=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin x-\cos x)dx-\int_{0}^{\frac{\pi}{4}}(\sin x-\cos x)dx \end{aligned}

\begin{aligned} &I=[-\cos x-\sin x]_{\frac{\pi}{4}}^{\frac{\pi}{2}}-|-\cos x-\sin x|_{0}^{\frac{\pi}{4}} \\\\ &I=-0-1+\sqrt{2}-(-\sqrt{2}+1)=2 \sqrt{2}-2 \\\\ &I=2(\sqrt{2}-1) \end{aligned}

 

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